## The image produced by a concave mirror is one- quarter the size of object is moved 5 cm closer to the mirror…..

Q.The image produced by a concave mirror is one- quarter the size of object is moved 5 cm closer to the mirror, the image will only be half the size of the object. The focal length of mirror is
(a) f = 5.0 cm

(b) f = 2.5 cm

(c) f = 7.5 cm

(d) f = 10cm

Ans: B

## A ray of light travelling on glass (μ = 3/2) is incident on a horizontal glass-air surface at the critical angle θc …..

Q.A ray of light travelling on glass (μ = 3/2) is incident on a horizontal glass-air surface at the critical angle θc. if a thin layer of water (μ = 4/3) is now poured on the glass-air surface, the angle at which the ray emerges into air the water-air surface is

(a) 60°
(b) 45°
(c) 90°
(d) 180°
Ans: C

## Light takes t1 sec to travel a distance x cm in vacuum and takes t2 sec to travel 10x cm in a medium. The critical angle…..

Q.Light takes t1 sec to travel a distance x cm in vacuum and takes t2 sec to travel 10x cm in a medium. The critical angle corresponding to the media is

(a)sin-1(10t1/t2)

(b) sin-1(t2/(10t1 ))

(c) sin-1(10t2/t1 )

(d) sin-1(t1/(10t2 ))

Ans:(a)

Sol: t1 = x/c , t2 = 10x/(c/μ ) = 10xμ/c

Hence , t2 = 10μt1 => μ= t2/10t1

sinθ = 1/μ =10t1/t2

θ = sin−1(10t1/t2)

## A rod of length 10cm lies along the principle axis of a concave mirror of focal length 10cm is such a way that …..

Q.A rod of length 10cm lies along the principle axis of a concave mirror of focal length 10cm is such a way that the end closer to the pole is 20cm away from it. The length of the image is

(a) 5cm

(b) 10cm

(c) 15cm

(d) 20cm

Ans: (a)

Sol: Radius of curvature R = 20 cm

Let Rod AB =10cm lies along principle axis in which end A is the centre . Hence image of A is at A itself .

For end B ,

1/v + 1/u = 1/f

1/v -1/(-30) = -1/10

1/v = 1/30 -1/10 =-2/30

v= -15 cm

Length of image = 5cm

## The angle of incidence on the surface of a diamond of refractive index 2.4, if the angle between the reflected and refracted rays is 90° is

Q.The angle of incidence on the surface of a diamond of refractive index 2.4, if the angle between the reflected and refracted rays is 90° is
(a) tan-1(2.4)
(b) sin-1(1/2.4)
(c) tan-1 (1/2.4)
(d) cos-1 (1/2.4)
Ans: A

Sol: According to question , i+r=90°  ….(i)

Applying Snell’s Law

1×sini = 2.4×sinr

sini = 2.4×sin(90-i) = 2.4cosi from(i)

tani = 2.4

i = tan−1(2.4)